# Even more Combinatorics Problems (Based on set {0,1,2,3,4,5,6}

• Nov 27th 2008, 06:59 PM
cnmath16
Even more Combinatorics Problems (Based on set {0,1,2,3,4,5,6}
28. a) How many 4-digit numbers can be formed from the set A = {0,1,2,3,4,5,6} if there is no repititon?

***Don't need help solving this... But it has to do with the other questions I need help with ***

c) How many of the numbers in part a) contain a 3?
* I know that numbers such as 0321 can't count because that actually represents the number 321. This is my thinking process thus far.

First digit (6 choices - excluding 0)
2nd digit (6 choices - can't be the 1st digit choice)
3rd digit (5 choices - can't be the 1st of the 2nd digit choice)
4th digit - I know I have to incorporate where the 3 is? I was thinking about adding it to the total?

If someone could show me how to solve this correctly that would be AWESOME!

Also need help with this... d) How many of the numbers in part a) are divisible by 5?

- The examples I have in my textbook refer to dividing many numbers at a time, (Using the Exclusion/Inclusion Principle) - which is why I am confused on what to do when there is just one number to divide.
• Nov 27th 2008, 08:33 PM
Soroban
Hello, cnmath16!

Quote:

a) How many 4-digit numbers can be formed from the set
$A \:=\: \{0,1,2,3,4,5,6\}$ if there is no repetiton?

The first digit cannot be 0: 6 choices.
Second digit: 6 choices.
Third digit: 5 choices.
Fourth digit: 4 choices.

There are: . $6\cdot6\cdot5\cdot4 \:=\:720$ possible four-digit numbers.

Quote:

c) How many of the numbers in part a) contain a 3?
How many numbers do not contain a 3?

. . The first digit cannot be 0 or 3: 5 choices.
. . Second digit: 5 choices.
. . Third digit: 4 choices.
. . Fourth digit: 3 choices

There are: . $5\cdot5\cdot4\cdot3 \:=\:300$ numbers without a 3.

Therefore, there are: . $720 - 300 \:=\:420$ numbers with a 3.

Quote:

d) How many of the numbers in part a) are divisible by 5?
There are two cases to consider:
. . (1) The number ends in 5.
. . (2) The number ends in 0.

(1) The number ends in 5: . _ _ _ 5

. . The first digit must not be 0 or 5: 5 choices.
. . Second digit: 5 choices.
. . Third digit: 4 choices.

There are: . $5\cdot5\cdot4 \:=\:100$ numbers ending in 5.

(2) The number ends in 0: ._ _ _ 0

. . First digit: 6 choices.
. . Second digit: 5 choices.
. . Third digit: 4 choices.

There are: . $6\cdot5\cdot4 \:=\:120$ numbers ending in 0.

Therefore, there are: . $100 + 120 \:=\:220$ numbers divisible by 5.